Question

question 3 of 10
which function does not have a vertical asymptote?
a. ( y = \frac{5x}{x + x^2} )
b. ( y = \frac{5x}{1 - 2x^2} )
c. ( y = \frac{5x - 1}{3 + x^2} )
d. ( y = \frac{x}{1 - x^2} )

Explanation:

Step1: Recall vertical asymptote condition

A vertical asymptote occurs where the denominator is zero (and numerator non - zero) for a rational function \(y = \frac{f(x)}{g(x)}\). So we need to check when the denominator of each function is zero.

Step2: Analyze Option A

For \(y=\frac{5x}{x + x^{2}}\), factor the denominator: \(x+x^{2}=x(1 + x)\). Set denominator equal to zero: \(x(1 + x)=0\). Solutions are \(x = 0\) and \(x=- 1\). So this function has vertical asymptotes at \(x = 0\) and \(x=-1\).

Step3: Analyze Option B

For \(y=\frac{5x}{1-2x^{2}}\), set denominator \(1 - 2x^{2}=0\). Then \(2x^{2}=1\), \(x^{2}=\frac{1}{2}\), \(x=\pm\frac{1}{\sqrt{2}}\). So this function has vertical asymptotes at \(x=\frac{1}{\sqrt{2}}\) and \(x =-\frac{1}{\sqrt{2}}\).

Step4: Analyze Option C

For \(y=\frac{5x - 1}{3+x^{2}}\), set denominator \(3+x^{2}=0\). But \(x^{2}=- 3\), and for real numbers, there is no solution (since the square of a real number is non - negative, \(x^{2}\geq0\), so \(x^{2}+3\geq3>0\) for all real \(x\)). So the denominator is never zero for real \(x\), so this function has no vertical asymptote.

Step5: Analyze Option D

For \(y=\frac{x}{1 - x^{2}}\), factor the denominator: \(1-x^{2}=(1 - x)(1 + x)\). Set denominator equal to zero: \((1 - x)(1 + x)=0\). Solutions are \(x = 1\) and \(x=-1\). So this function has vertical asymptotes at \(x = 1\) and \(x=-1\).

Answer:

C. \(y=\frac{5x - 1}{3+x^{2}}\)